Spectroscopy & Analysis (Intro)

Question 1 (10)

(a) Derivation (6)

• Consider a slab of thickness $dx$ at depth $x$; incident intensity $I$. The fractional drop $-\dfrac{dI}{I}$ is proportional to number of absorbers $\propto c\,dx$: (1) $-\frac{dI}{I} = k\,c\,dx$ (1)
• Integrate from $0$ to $l$ ($I_0 \to I$): (1) $-\int_{I_0}^{I}\frac{dI}{I} = kc\int_0^l dx \Rightarrow \ln\frac{I_0}{I} = kcl$ (1)
• Convert to base-10 ($\ln = 2.303\log$), absorbing constant into $\varepsilon$: (1) $A = \log\frac{I_0}{I} = \varepsilon c l$ where $\varepsilon = k/2.303$. (1)

(b) (3)

• $A = -\log T = -\log(0.25) = 0.602$ (1)
• $c = A/(\varepsilon l) = 0.602/(1.20\times10^4 \times 1.00)$ (1)
• $c = 5.02\times10^{-5}\ \text{mol L}^{-1}$ (1)

(c) (1) At high $c$: intermolecular interactions / change in refractive index / analyte aggregation alter $\varepsilon$; or non-monochromatic light — any one. (1)

Question 2 (8)

(a) (3) $E_{photon} = hc/\lambda$ (1); per mole multiply by $N_A$ (1): $E = \frac{N_A hc}{\lambda} = \frac{6.022\times10^{23}\times6.626\times10^{-34}\times3.00\times10^8}{\lambda[\text{m}]}\ \text{J mol}^{-1} = \frac{1.196\times10^{5}}{\lambda[\text{nm}]}\ \text{kJ mol}^{-1}$ (1)

(b) (2) $E = 1.196\times10^5/300 = 399\ \text{kJ mol}^{-1}$ (≈$399$). (2)

(c) (3) Increasing energy: NMR (nuclear spin flip) < IR (bond vibration) < UV-Vis (electronic transition) < X-ray (core-electron / inner-shell ionisation). (3) (½ each ordering + property, rounded to 3)

Question 3 (8)

(a) (4) $\tilde\nu \propto \sqrt{k/\mu}$. O–H has high force constant (strong bond) and small reduced mass ($\mu$ dominated by light H) → high $\tilde\nu$ (~3600) (1); C=O double bond has large $k$ but heavier atoms, and lower than O–H → ~1715 (1). For C–D vs C–H: force constant $k$ nearly unchanged, but $\mu$ larger for D → $\tilde\nu$ lower (2) (factor ≈$1/\sqrt2$).

(b) (2) Aldehyde shows the C–H stretch of the aldehyde group near $2720$ and $2820\ \text{cm}^{-1}$ (Fermi doublet), absent in the ketone. (2)

(c) (2) Fingerprint region ≈ $1500$–$400\ \text{cm}^{-1}$: complex pattern of skeletal/bending vibrations unique to a molecule (1); used to confirm identity by matching against a reference spectrum even without assigning individual bands (1).

Question 4 (12)

(a) (5) Each of $n$ equivalent neighbouring protons is independently spin-up ($\alpha$) or spin-down ($\beta$) (1); these combine to give the local field felt by the observed proton (1). Number of distinguishable total spin states = $n+1$ lines (1). For $n=3$: multiplicities of arrangements are $\alpha\alpha\alpha$ (1), 3× (one $\beta$), 3× (two $\beta$), $\beta\beta\beta$ (1) → 1:3:3:1 quartet (1), matching binomial coefficients (Pascal's triangle) (1).

(b) (5)

• $\text{CH}_3$ (of $\text{COCH}_3$): singlet, ~$\delta 2.0$, 3H (no neighbouring H) (1)
• $\text{OCH}_2$: quartet, ~$\delta 4.1$, 2H (3 neighbours) (1)
• $\text{CH}_3$ (of ethyl): triplet, ~$\delta 1.2$, 3H (2 neighbours) (1)
• 3 signals total (1); integration ratio $3:2:3$ (1)

(c) (2) e.g. dimethyl ether $\text{CH}_3\text{OCH}_3$ — all 6 H equivalent → one singlet; (ethanol would give three signals). (2) (Accept any molecule with a single H environment justified.)

Question 5 (10)

(a) (5) $M=74$ with loss of 15 (=$\text{CH}_3$) and $m/z=31$ (=$\text{CH}_2\text{OH}^+$ or $\text{CH}_3\text{O}^+$). Structure: methyl propanoate or better a molecule like $\text{C}_3\text{H}_6\text{O}_2$... Best fit: 1-propanol? (M=60) no. $M=74 = \text{C}_4\text{H}_{10}\text{O}$ (e.g. butan-1-ol, or diethyl ether). Loss of 15 = $\text{CH}_3$ (1); $m/z=31 = \text{CH}_2=\overset{+}{\text O}\text H$ (characteristic of primary alcohol) (1). Propose butan-2-ol / 2-methylpropan-2-ol giving $M-15=59$; the $m/z=31$ oxocarbenium confirms O present (1). Award marks for: correct molecular formula $\text{C}_4\text{H}_{10}\text{O}$ (1), correct fragment identities and mechanism (α-cleavage) (2).

(b) (2) Roughly equal $M$ and $M+2$ → bromine present (isotopes $^{79}\text{Br}:^{81}\text{Br} \approx 1:1$). (2) (Chlorine would give ~3:1.)

(c) (3) $M^+$ of $\text{C}_2\text{H}_5{}^{79}\text{Br} = 24+5+79 = 108$ (1.5); after loss of Br ($-79$): $\text{C}_2\text{H}_5^+ = 29$ (1.5).

Question 6 (12)

(a) (4) TLC separates by differential adsorption/partition between a polar stationary phase (silica) and a moving solvent (mobile phase): compounds with greater affinity for the mobile phase / less for stationary travel further (2). $R_f = \dfrac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}}$ (1). $R_f = 3.6/8.0 = 0.45$ (1).

(b) (2) GC: separation by volatility / boiling point (and interaction with stationary phase) in the gas phase (1). HPLC: separation by polarity / partition between liquid mobile phase and stationary phase under high pressure (1).

(c) (6) Degrees of unsaturation of $\text{C}_4\text{H}_8\text{O}_2 = (2\cdot4+2-8)/2 = 1$ → one C=O (1). IR $1740$, no broad O–H → ester (not acid) (1). NMR: triplet 3H + quartet 2H = ethyl group $\text{OCH}_2\text{CH}_3$ ($\delta4.1$ shows O-attached CH₂) (1); singlet 3H at $\delta2.0$ = $\text{CH}_3$ adjacent to C=O, isolated (1). Assemble: ethyl ethanoate (ethyl acetate) $\text{CH}_3\text{COOCH}_2\text{CH}_3$ (1); check mass $= 88$ ✓ (1).

[
  {"claim":"Beer-Lambert: A=-log10(0.25), c=A/(1.2e4*1)=5.02e-5","code":"A=-log(Rational(1,4),10); c=A/(Integer(12000)*1); result = abs(float(c)-5.02e-5)<1e-6"},
  {"claim":"Photon energy per mole at 300nm ~399 kJ/mol","code":"E=(6.022e23*6.626e-34*3.00e8)/(300e-9)/1000; result = abs(float(E)-399)<2"},
  {"claim":"Rf = 3.6/8.0 = 0.45","code":"result = abs(float(Rational(36,80))-0.45)<1e-9"},
  {"claim":"Bromoethane M+ (79Br)=108 and C2H5+ =29","code":"M=24+5+79; frag=24+5; result = (M==108) and (frag==29)"},
  {"claim":"Degrees of unsaturation C4H8O2 = 1","code":"DoU=(2*4+2-8)/2; result = DoU==1"}
]